Is this proof correct?
Proof: Here $\theta = \arccos(6/7)$. Now to show we can't trisect $\theta$, we show that $\theta/3$ is not constructible by finding the irreducible polynomial in $\mathbb Q[x]$, whose root it is. By the angle tripling formula we have that, $\cos(3\theta)=4\cos^3(\theta)-3\cos(\theta) \Rightarrow $ $\cos(\theta)=4\cos^3(\theta/3)-3\cos(\theta/3)$. Here we see that $\cos(\theta)=6/7$. Thus we have, $6/7=4\cos^3(\theta/3)-3\cos(\theta/3)$. Take $x=\cos(\theta/3)$. So now we have, $28x^3-21x-6=0 \Rightarrow f(x)=28x^3-21x-6$ is a polynomial with root $\cos(\theta/3)$. But $f(x)$ is irreducible by Eisensteins Criterion with $p=3$, thus $[\mathbb Q[\cos(\theta/3)]: \mathbb Q]=3$ which is not a power of two.